Answer
After the collision, the 100-g ball is moving in the -x-direction at a speed of 2.50 m/s
After the collision, the 300-g ball is moving in the +x-direction at a speed of 2.50 m/s
Work Step by Step
Let $m_A = 0.10~kg$ and let $m_B = 0.30~kg$
Let $v_A$ be the initial velocity of the 100-g ball.
Let $v_B$ be the initial velocity of the 300-g ball.
Let $v_A'$ be the final velocity of the 100-g ball.
Let $v_B'$ be the final velocity of the 300-g ball.
We can use conservation of momentum to set up an equation.
$m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can set up another equation.
$v_A - v_B = v_B' - v_A'$
$v_A' = v_B' - v_A + v_B$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_B’ = \frac{(2)(0.10~kg)(5.00~m/s)+(0.40~kg)(0)- (0.10~kg)(0)}{(0.10~kg)+(0.30~kg)}$
$v_B' = 2.50~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = 2.50~m/s - 5.00~m/s + 0$
$v_A' = -2.50~m/s$
After the collision, the 100-g ball is moving in the -x-direction at a speed of 2.50 m/s
After the collision, the 300-g ball is moving in the +x-direction at a speed of 2.50 m/s